Explanation:
Since, the rod is present in vertical position and the spring is unrestrained.
So, initial potential energy stored in the spring is
= 0
And, initial potential gravitational potential energy of the rod is
.
It is given that,
mass of the bar = 0.795 kg
g = 9.8 ![m/s^{2}](https://tex.z-dn.net/?f=m%2Fs%5E%7B2%7D)
L = length of the rod = 0.2 m
Initial total energy T = ![\frac{mgL}{2}](https://tex.z-dn.net/?f=%5Cfrac%7BmgL%7D%7B2%7D)
Now, when the rod is in horizontal position then final total energy will be as follows.
T = ![\frac{1}{2}kx^{2} + I \omega^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dkx%5E%7B2%7D%20%2B%20I%20%5Comega%5E%7B2%7D)
where, I = moment of inertia of the rod about the end = ![\frac{mL^{2}}{3}](https://tex.z-dn.net/?f=%5Cfrac%7BmL%5E%7B2%7D%7D%7B3%7D)
Also, ![\omega = \frac{\nu}{L}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Cfrac%7B%5Cnu%7D%7BL%7D)
where,
= speed of the tip of the rod
x = spring extension
The initial unstrained length is
= 0.1 m
Therefore, final length will be calculated as follows.
x' =
m
Then, x = ![x' - x_{o}](https://tex.z-dn.net/?f=x%27%20-%20x_%7Bo%7D)
x =
m - 0.1 m
= 0.1236 m
k = 25 N/m
So, according to the law of conservation of energy
![\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}](https://tex.z-dn.net/?f=%5Cfrac%7BmgL%7D%7B2%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7Dkx%5E%7B2%7D%20%2B%20%5Cfrac%7B1%20%5Ctimes%20mL%5E%7B2%7D%7D%7B2%20%5Ctimes%203%7D%28%5Cfrac%7B%5Cnu%7D%7BL%7D%29%5E%7B2%7D)
![\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}](https://tex.z-dn.net/?f=%5Cfrac%7BmgL%7D%7B2%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7Dkx%5E%7B2%7D%20%2B%20%5Cfrac%7B1%7D%7B6%7Dmv%5E%7B2%7D)
Putting the given values into the above formula as follows.
![\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}](https://tex.z-dn.net/?f=%5Cfrac%7BmgL%7D%7B2%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7Dkx%5E%7B2%7D%20%2B%20%5Cfrac%7B1%7D%7B6%7Dmv%5E%7B2%7D)
![\frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B0.795%20kg%20%5Ctimes%209.8%20%5Ctimes%200.2%20m%7D%7B2%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%2027%20N%2Fm%20%5Ctimes%20%280.1236%29%5E%7B2%7D%20%2B%20%5Cfrac%7B1%7D%7B6%7D%20%5Ctimes%200.795%20%5Ctimes%20v%5E%7B2%7D)
v = 2.079 m/s
Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.